標題:
F.4 PHYSICS
發問:
A horizontal force F is applied to a block of mass M on a rough horizontal surface. The acceleration of the block is a. If the force is changed to 2F and the frictional force remains unchanged,then the accelearation of the block will be greater then 2a. Explain
最佳解答:
When a force of F is applied to the block, F - Ff = M.a where Ff is the frcitional force. If the force increases to 2F, then, 2F - Ff = Ma' where a' is the neww acceleration of the block. Hence, by dividing the two equations, Ma'/Ma = (2F - Ff)/(F-Ff) i.e. a'/a = (2F - Ff)/(F-Ff) a'/a = 2(F - Ff/2)/(F-Ff) But (F-Ff/2) > (F-Ff) i.e. (F - Ff/2)/(F-Ff) > 1 hence, a'/a > 2 or a' > 2a
F.4 PHYSICS
發問:
A horizontal force F is applied to a block of mass M on a rough horizontal surface. The acceleration of the block is a. If the force is changed to 2F and the frictional force remains unchanged,then the accelearation of the block will be greater then 2a. Explain
最佳解答:
When a force of F is applied to the block, F - Ff = M.a where Ff is the frcitional force. If the force increases to 2F, then, 2F - Ff = Ma' where a' is the neww acceleration of the block. Hence, by dividing the two equations, Ma'/Ma = (2F - Ff)/(F-Ff) i.e. a'/a = (2F - Ff)/(F-Ff) a'/a = 2(F - Ff/2)/(F-Ff) But (F-Ff/2) > (F-Ff) i.e. (F - Ff/2)/(F-Ff) > 1 hence, a'/a > 2 or a' > 2a
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